वर्गमूल सर्पिल में \(\sqrt{7}\) को दर्शाने वाला खंड किस प्रकार मिलता है?
How is the segment representing \(\sqrt{7}\) obtained in a square root spiral?
Explanation opens after your attempt
Correct Answer
B. \(\sqrt{6}\) पर (1) इकाई लंब बनाकरBy drawing (1) unit perpendicular on \(\sqrt{6}\)
Concept
Because (\(\sqrt{6}\)2+12=7), the new segment becomes \(\sqrt{7}\). Identifying the correct previous segment is necessary.
Why this answer is correct
The correct answer is B. \(\sqrt{6}\) पर (1) इकाई लंब बनाकर / By drawing (1) unit perpendicular on \(\sqrt{6}\). Because (\(\sqrt{6}\)2+12=7), the new segment becomes \(\sqrt{7}\). Identifying the correct previous segment is necessary.
Exam Tip
क्योंकि (\(\sqrt{6}\)2+12=7), इसलिए नया खंड \(\sqrt{7}\) होगा। सही पिछले खंड को पहचानना जरूरी है।
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